This is (mostly) a pretty routine optimization problem. The methods of (for example) a standard calculus class are enough to tell you that $a+b+c$ will be largest when two of $a,b,c$ are as small as possible and the third is whatever it has to be. So if you don't care whether the variables are integers, take $a=1,b=2,c=(n-2)/3$.

Thus if $n$ has the form $3k+2$, the optimum is achieved in integers. Take $a=1,b=2,c=k$, and then we have $a+b+c=3+\frac{n-2}{3}=\frac{n+7}{3}$.

So $\frac{n+7}{3}$ is an upper bound, and it actually gives the correct answer infinitely often.